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3 years ago

Find the m < t rounded to the nearest degree

Mathematics

1 answer:

serg [7]3 years ago

5 0

Answer:

28° (Answer A)

Step-by-step explanation:

Note that tan t = 8/15 = 0.533

Applying the inverse tangent function, we get

t = arctan 0.533 = 0.4900 radian

Mult. this by the radian-degree conversion factor results in: 28.072°

and when rounded off to the nearest degree, the measure of angle T is 28° (Answer A)

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Each time Kristine gets paid, she spends $20 and saves the rest. If the amount Kristine earns is represented by x and the amount

faust18 [17]

We are given that Kristine spends $20 and saves the rest each time she get paid.

We can use slope-intercept form y=mx+b to represent the equation.

Where x represents the amount Kristine earns and y represents the amount she saves.

Kristine spends $20. Therefore b= -20.

Plugging mx as just x and b=-20.

<h3>y = x-20.</h3><h3>If we plug y=0, we get </h3><h3>0 = x-20</h3><h3>x=20.</h3><h3>We can see in 4th option we have x-intercept =20.</h3><h3>Therefore, correct option is 4th option. </h3><h3 />

4 0

3 years ago

Read 2 more answers

Given sec theta=-4/3 and 90°

hoa [83]

Answer:

Option A.

Step-by-step explanation:

It is given secθ = - $\frac{4}{3}$

then cosθ = - $\frac{3}{4}$

Now we know the identity

sinθ = $\sqrt{1-cos^{2}\theta }$

= $\sqrt{1-(-\frac{3}{4} )^{2}}$

= $\frac{\sqrt{7} }{4}$

Now sin2θ = 2sinθcosθ

Now we put the values of cosine and sine in the formula

sin2θ = 2×(- $\frac{3}{4}$ )( $\frac{\sqrt{7}}{4}$ )

= - $\frac{3\sqrt{7}}{8}$

Therefore, option A. is the answer.

3 0

3 years ago

Which of the following expressions is a factor of the polynomial x 2 +3/2x-1

Elodia [21]

Answer:

$\large \boxed{\sf \ \ \ (x+2)(x-\dfrac{1}{2}) \ \ \ }$

Step-by-step explanation:

Hello,

let's solve

$x^2+\dfrac{3}{2}x-1=0\\\\ 2x^2+3x-2=0 \ \text{multiply by 2}\\\\\\$

$\Delta=b^2-4ac=9+4*2*2=9+16=25$

There are two solutions

$x_1=\dfrac{-3-\sqrt{25}}{4}\\\\x_1=\dfrac{-3-5}{4}\\\\x_1=\dfrac{-8}{4}\\\\\boxed{x_1=-2}$

And

$x_2=\dfrac{-3+\sqrt{25}}{4}\\\\x_2=\dfrac{-3+5}{4}\\\\x_2=\dfrac{2}{4}\\\\\boxed{x_2=\dfrac{1}{2}}$

So we can write

$x^2+\dfrac{3}{2}x-1\\\\=(x-x_1)(x-x_2)\\\\=(x+2)(x-\dfrac{1}{2})$

Hope this helps.

Do not hesitate if you need further explanation.

Thank you

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3 years ago

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podryga [215]

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